JEE Main 2025 — 29 January, Morning Shift — Question 68

(i) Single letter is used , then no. of words $=5$

(ii) Two distinct letters are used, then no. of words${ }^{5} \mathrm{C}_{2} \times\left(\frac{6!}{2!4!} \times 2+\frac{6!}{3!3!}\right)=10(30+20)=500$

(iii) Three distinct letters are used, then no. of words

${ }^{5} \mathrm{C}_{3} \times \frac{6!}{2!2!2!}=900$

Total no. of words $=1405$